Answer
(a) $2.6\times 10^{-4}C$
(b) $28mA$
Work Step by Step
(a) We can find the required charge on the capacitor as
$q=C\mathcal{E}(1-e^{\frac{-t}{t}})$
$q=C\mathcal{E}(1-e^{-1})$
We plug in the known values to obtain:
$q=(45\times 10^{-6}F)(9.0V)(1-0.37)$
$q=2.6\times 10^{-4}C $
(b) We can find the current in the circuit as
$I=(\frac{\mathcal{E}}{R})e^{\frac{-t}{t}}$
$I=(\frac{\mathcal{E}}{R})e^{-1}$
We plug in the known values to obtain:
$I=(\frac{9.0V}{120\Omega})(0.37)$
$I=0.02775A$
$I=28mA$
